All groups of order 14

Let G represent the order 14 group. Claim is either ${\displaystyle G\stackrel{\sim }{=}{\mathbb{Z}}_{14}}$, i.e., the cyclic group of order 14 or ${\displaystyle G\stackrel{\sim }{=}{D}_{14}}$, i.e., the dihedral group of order 14. It is well known fact that a group of even order must contain an element of order 2, say a. Now, we are aware that the order of an element in a finite group must be divided by the Lagrange theorem. Therefore, the order will be one of 1, 2, 7, or 14. We will use the symbol b to indicate the element of order 7 that we want to demonstrate.
case-1: Suppose there exist an element order of $c=14$, So, we take ${\displaystyle c^2=b}$, therefore order of b is 7 and cc generates G, hence G is a cyclic group of order 14.
case-2: Assume that neither an element of order 14 nor 7 exists. Therefore, every element except for a,e will have order 2. Let c now differ from a, which is a component of order 2. Now, according to our presumption, AC is a member of order 2. It is demonstrable that $\{e,a,c,ac\}$ forms a subgroup of G, (since $ac=ca$). But this is impossible, since order of subgroup divides order of G. So, there exists an element of order 7, say b. Now consider ${\displaystyle H=\{e,b,b^2,\dots ,b^6\}}$ is subgroup of order 7. Since aa has order 2 so, aa not in H, because all the elements of H are of order 7. Therefore, H and aHaH two distinct disjoint cosets of H, each of them consists of 7 elements. Therefore,
${\displaystyle G=\{e,b,b^2,\dots ,a,ab,a{b}^2,\dots a{b}^6\}.}$
Now consider the element ba, it must be one of the ab^j, where ${\displaystyle j\in \{1,2,\cdots \hspace{0.17em},6\}}$. Suppose $ba=ab$, then G will be abelian, since G is collection of elements like ${\displaystyle a^i{b}^k}$, where ${\displaystyle 0\le i\le 1}$ and ${\displaystyle 0\le k\le 6}$ and order of abab will be 14, hence G will be cyclic. Otherwise, Since by sylow theorem ${\displaystyle GG}$ have only one subgroup of order 7 and that is H, so, order of baba must be 2. ( Order of baba can not be 7, since baba not in H and order of baba can not be 14, otherwise G is cyclic and hence abelian, which is not the case.).Therefore,
${\displaystyle ba={\left(ba\right)}^{-1}=a^{-1}{b}^{-1}-a{b}^6(since{a}^2=b^7=e)}$
Therefore,
${\displaystyle G=\{a^i{b}^k:where0\le i\le 1\text{and}0\le k\le 6,ba=a{b}^6\}}$, that is it is isomorphic to ${\displaystyle D_{14}}$